Properties & Defintions of the First Uncountable Ordinal
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I am confused about the concept of the “first uncountable ordinal.” Here are 3 statements that are apparently equivalent:
(1) There is an uncountable ordinal that is “first” in the sense that it is the smallest ordinal that, when considered as a set, is uncountable.
(2) This ordinal is the supremum of all countable ordinals.
(3) The elements of the set are the countable ordinals, of which there are uncountable many.
First, which statement is generally given as the definition of “first uncountable ordinal”?
Secondly, how do we know the supremum in statement (2) exists?
Third, why does statement (3) follow from the other statements? And vice versa?
set-theory
edited Jan 25 at 1:19
Andrés E. Caicedo
65.6k8160250
asked Jan 24 at 21:47
MPittsMPitts
739412
$endgroup$
add a comment |
$begingroup$
I am confused about the concept of the “first uncountable ordinal.” Here are 3 statements that are apparently equivalent:
(1) There is an uncountable ordinal that is “first” in the sense that it is the smallest ordinal that, when considered as a set, is uncountable.
(2) This ordinal is the supremum of all countable ordinals.
(3) The elements of the set are the countable ordinals, of which there are uncountable many.
First, which statement is generally given as the definition of “first uncountable ordinal”?
Secondly, how do we know the supremum in statement (2) exists?
Third, why does statement (3) follow from the other statements? And vice versa?
set-theory
edited Jan 25 at 1:19
Andrés E. Caicedo
65.6k8160250
asked Jan 24 at 21:47
MPittsMPitts
739412
$endgroup$
1
$begingroup$
en.wikipedia.org/wiki/Hartogs_number $leftarrow$ Use this with $X=omega$.
$endgroup$
– J.G.
Jan 24 at 22:10
$begingroup$
(3) Is just a consequence of the particular way we've defined ordinals and (2). Every ordinal is the set of all ordinals below it, and being a supremum of countable ordinals means the ones below it are exactly the countable ones. I leave this as a comment because I think it's the least substantial of the three.
$endgroup$
– Malice Vidrine
Jan 24 at 23:45
add a comment |
$begingroup$
I am confused about the concept of the “first uncountable ordinal.” Here are 3 statements that are apparently equivalent:
(1) There is an uncountable ordinal that is “first” in the sense that it is the smallest ordinal that, when considered as a set, is uncountable.
(2) This ordinal is the supremum of all countable ordinals.
(3) The elements of the set are the countable ordinals, of which there are uncountable many.
First, which statement is generally given as the definition of “first uncountable ordinal”?
Secondly, how do we know the supremum in statement (2) exists?
Third, why does statement (3) follow from the other statements? And vice versa?
set-theory
edited Jan 25 at 1:19
Andrés E. Caicedo
65.6k8160250
asked Jan 24 at 21:47
MPittsMPitts
739412
$endgroup$
I am confused about the concept of the “first uncountable ordinal.” Here are 3 statements that are apparently equivalent:
(1) There is an uncountable ordinal that is “first” in the sense that it is the smallest ordinal that, when considered as a set, is uncountable.
(2) This ordinal is the supremum of all countable ordinals.
(3) The elements of the set are the countable ordinals, of which there are uncountable many.
First, which statement is generally given as the definition of “first uncountable ordinal”?
Secondly, how do we know the supremum in statement (2) exists?
Third, why does statement (3) follow from the other statements? And vice versa?
set-theory
set-theory
edited Jan 25 at 1:19
Andrés E. Caicedo
65.6k8160250
asked Jan 24 at 21:47
MPittsMPitts
739412
edited Jan 25 at 1:19
Andrés E. Caicedo
65.6k8160250
asked Jan 24 at 21:47
MPittsMPitts
739412
edited Jan 25 at 1:19
Andrés E. Caicedo
65.6k8160250
edited Jan 25 at 1:19
Andrés E. Caicedo
65.6k8160250
edited Jan 25 at 1:19
Andrés E. Caicedo
65.6k8160250
65.6k8160250
asked Jan 24 at 21:47
MPittsMPitts
739412
asked Jan 24 at 21:47
MPittsMPitts
739412
asked Jan 24 at 21:47
MPittsMPitts
739412
739412
1
$begingroup$
en.wikipedia.org/wiki/Hartogs_number $leftarrow$ Use this with $X=omega$.
$endgroup$
– J.G.
Jan 24 at 22:10
$begingroup$
(3) Is just a consequence of the particular way we've defined ordinals and (2). Every ordinal is the set of all ordinals below it, and being a supremum of countable ordinals means the ones below it are exactly the countable ones. I leave this as a comment because I think it's the least substantial of the three.
$endgroup$
– Malice Vidrine
Jan 24 at 23:45
add a comment |
1
$begingroup$
en.wikipedia.org/wiki/Hartogs_number $leftarrow$ Use this with $X=omega$.
$endgroup$
– J.G.
Jan 24 at 22:10
$begingroup$
(3) Is just a consequence of the particular way we've defined ordinals and (2). Every ordinal is the set of all ordinals below it, and being a supremum of countable ordinals means the ones below it are exactly the countable ones. I leave this as a comment because I think it's the least substantial of the three.
$endgroup$
– Malice Vidrine
Jan 24 at 23:45
1
1
$begingroup$
en.wikipedia.org/wiki/Hartogs_number $leftarrow$ Use this with $X=omega$.
$endgroup$
– J.G.
Jan 24 at 22:10
$begingroup$
en.wikipedia.org/wiki/Hartogs_number $leftarrow$ Use this with $X=omega$.
$endgroup$
– J.G.
Jan 24 at 22:10
$begingroup$
(3) Is just a consequence of the particular way we've defined ordinals and (2). Every ordinal is the set of all ordinals below it, and being a supremum of countable ordinals means the ones below it are exactly the countable ones. I leave this as a comment because I think it's the least substantial of the three.
$endgroup$
– Malice Vidrine
Jan 24 at 23:45
$begingroup$
(3) Is just a consequence of the particular way we've defined ordinals and (2). Every ordinal is the set of all ordinals below it, and being a supremum of countable ordinals means the ones below it are exactly the countable ones. I leave this as a comment because I think it's the least substantial of the three.
$endgroup$
– Malice Vidrine
Jan 24 at 23:45
add a comment |
1 Answer
1
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votes
$begingroup$
(1) is the definition.
Since the ordinals are well ordered, one just needs to show there is an uncountable ordinal.
Use Hartogs lemma. Alternatively, use the axiom of choice to well order the reals.
Let u be the first uncountable ordinal.
If k is a countable ordinal, k < u.
So u is an upper bound of countable ordinals.
If d is an upper bound of the countable ordinals and d < u, then d is countable. As d + 1 is countable, d + 1 <= d, a contradiction.
So u is the supremum of the countable ordinals.
answered Jan 24 at 22:49
William ElliotWilliam Elliot
8,5922720
$endgroup$
$begingroup$
We don't need AC . Let $B$ be the set of subsets of $omega times omega$ that are (graphs of) well-orderings. Each $bin B$ is isomorphic to a $unique$ (countable) ordinal $F(b).$ Then $ {F(b): bin B} =omega_1$.
$endgroup$
– DanielWainfleet
Jan 26 at 16:58
add a comment |
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$begingroup$
(1) is the definition.
Since the ordinals are well ordered, one just needs to show there is an uncountable ordinal.
Use Hartogs lemma. Alternatively, use the axiom of choice to well order the reals.
Let u be the first uncountable ordinal.
If k is a countable ordinal, k < u.
So u is an upper bound of countable ordinals.
If d is an upper bound of the countable ordinals and d < u, then d is countable. As d + 1 is countable, d + 1 <= d, a contradiction.
So u is the supremum of the countable ordinals.
answered Jan 24 at 22:49
William ElliotWilliam Elliot
8,5922720
$endgroup$
$begingroup$
We don't need AC . Let $B$ be the set of subsets of $omega times omega$ that are (graphs of) well-orderings. Each $bin B$ is isomorphic to a $unique$ (countable) ordinal $F(b).$ Then $ {F(b): bin B} =omega_1$.
$endgroup$
– DanielWainfleet
Jan 26 at 16:58
add a comment |
$begingroup$
(1) is the definition.
Since the ordinals are well ordered, one just needs to show there is an uncountable ordinal.
Use Hartogs lemma. Alternatively, use the axiom of choice to well order the reals.
Let u be the first uncountable ordinal.
If k is a countable ordinal, k < u.
So u is an upper bound of countable ordinals.
If d is an upper bound of the countable ordinals and d < u, then d is countable. As d + 1 is countable, d + 1 <= d, a contradiction.
So u is the supremum of the countable ordinals.
answered Jan 24 at 22:49
William ElliotWilliam Elliot
8,5922720
$endgroup$
$begingroup$
We don't need AC . Let $B$ be the set of subsets of $omega times omega$ that are (graphs of) well-orderings. Each $bin B$ is isomorphic to a $unique$ (countable) ordinal $F(b).$ Then $ {F(b): bin B} =omega_1$.
$endgroup$
– DanielWainfleet
Jan 26 at 16:58
add a comment |
$begingroup$
(1) is the definition.
Since the ordinals are well ordered, one just needs to show there is an uncountable ordinal.
Use Hartogs lemma. Alternatively, use the axiom of choice to well order the reals.
Let u be the first uncountable ordinal.
If k is a countable ordinal, k < u.
So u is an upper bound of countable ordinals.
If d is an upper bound of the countable ordinals and d < u, then d is countable. As d + 1 is countable, d + 1 <= d, a contradiction.
So u is the supremum of the countable ordinals.
answered Jan 24 at 22:49
William ElliotWilliam Elliot
8,5922720
$endgroup$
(1) is the definition.
Since the ordinals are well ordered, one just needs to show there is an uncountable ordinal.
Use Hartogs lemma. Alternatively, use the axiom of choice to well order the reals.
Let u be the first uncountable ordinal.
If k is a countable ordinal, k < u.
So u is an upper bound of countable ordinals.
If d is an upper bound of the countable ordinals and d < u, then d is countable. As d + 1 is countable, d + 1 <= d, a contradiction.
So u is the supremum of the countable ordinals.
answered Jan 24 at 22:49
William ElliotWilliam Elliot
8,5922720
answered Jan 24 at 22:49
William ElliotWilliam Elliot
8,5922720
answered Jan 24 at 22:49
William ElliotWilliam Elliot
8,5922720
answered Jan 24 at 22:49
William ElliotWilliam Elliot
8,5922720
8,5922720
$begingroup$
We don't need AC . Let $B$ be the set of subsets of $omega times omega$ that are (graphs of) well-orderings. Each $bin B$ is isomorphic to a $unique$ (countable) ordinal $F(b).$ Then $ {F(b): bin B} =omega_1$.
$endgroup$
– DanielWainfleet
Jan 26 at 16:58
add a comment |
$begingroup$
We don't need AC . Let $B$ be the set of subsets of $omega times omega$ that are (graphs of) well-orderings. Each $bin B$ is isomorphic to a $unique$ (countable) ordinal $F(b).$ Then $ {F(b): bin B} =omega_1$.
$endgroup$
– DanielWainfleet
Jan 26 at 16:58
$begingroup$
We don't need AC . Let $B$ be the set of subsets of $omega times omega$ that are (graphs of) well-orderings. Each $bin B$ is isomorphic to a $unique$ (countable) ordinal $F(b).$ Then $ {F(b): bin B} =omega_1$.
$endgroup$
– DanielWainfleet
Jan 26 at 16:58
$begingroup$
We don't need AC . Let $B$ be the set of subsets of $omega times omega$ that are (graphs of) well-orderings. Each $bin B$ is isomorphic to a $unique$ (countable) ordinal $F(b).$ Then $ {F(b): bin B} =omega_1$.
$endgroup$
– DanielWainfleet
Jan 26 at 16:58
add a comment |
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1
$begingroup$
en.wikipedia.org/wiki/Hartogs_number $leftarrow$ Use this with $X=omega$.
$endgroup$
– J.G.
Jan 24 at 22:10
$begingroup$
(3) Is just a consequence of the particular way we've defined ordinals and (2). Every ordinal is the set of all ordinals below it, and being a supremum of countable ordinals means the ones below it are exactly the countable ones. I leave this as a comment because I think it's the least substantial of the three.
$endgroup$
– Malice Vidrine
Jan 24 at 23:45